CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 1 PART A Energetics in Chemical Reaction USEFUL CONCEPTS AND FORMULAS 1. )G o (reaction) = 3 n )G o f (products) ! 3 n )G o f (reactants) )G o < 0, the forward reaction occurs spontaneous. )G o = 0, the reaction system is at equilibrium. )G o > 0, the forward reaction is nonspontaneous. 2. )G = )H ! T )S )G depends on T, but )H and )S dont. For a phase transition, )G = 0 3. )S (reaction) = 3 n )S (products) ! 3 n )S (reactants) )S > 0, more random (disorder) in products than in reactants S gas > S liquid > S solid 4. Effect of Temperature on the Spontaneity of Reaction )H )S Temperature )G = )H ! T )S Spontaneous direction s r all always s forward s s low s forward high r reverse r r low r reverse high s forward r s all always r reverse
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PART A Energetics in Chemical ReactionCHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 1 PART A Energetics in Chemical Reaction USEFUL CONCEPTS AND FORMULAS 1.)Go (reaction)
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CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 1
PART A Energetics in Chemical Reaction
USEFUL CONCEPTS AND FORMULAS
1.)Go (reaction) = 3 n )Go
f (products) ! 3 n )Gof (reactants)
)Go < 0, the forward reaction occurs spontaneous. )Go = 0, the reaction system is at equilibrium. )Go > 0, the forward reaction is nonspontaneous.
2.)G = )H ! T )S )G depends on T, but )H and )S don�t.
For a phase transition, )G = 0
3.)S (reaction) = 3 n )S (products) ! 3 n )S (reactants)
)S > 0, more random (disorder) in products than in reactantsSgas > Sliquid > Ssolid
4. Effect of Temperature on the Spontaneity of Reaction
)H )S Temperature )G = )H ! T )S Spontaneous direction
s r all always s forward
s s low s forwardhigh r reverse
r r low r reversehigh s forward
r s all always r reverse
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 2
Compare to a straight-line linear equation:y = m x + b
where y = Rn Keqx = 1 / Tm = slope = ! )Ho / Rb = y-intercept = )So / R
5. Under non-standard state conditions,
)G = )Go + R T Rn Q
6. At equilibrium, Q = Keq and )G = 0OR
)Go = ! R T Rn Keq
Keq = Kc for aqueous reactants and products in terms of mol L!1 (M) Keq = Kp for gaseous reactants and products in terms of atm
)Go < 0, Rn Keq > 0, Keq > 1 the products are favoured over reactants(i.e. the forward reaction proceeds spontaneously)
)Go = 0, Rn Keq = 0, Keq = 1 the reaction is at equilibrium )Go > 0, Rn Keq < 0, Keq < 1 the reactants are favoured over products
(i.e. the reverse reaction is spontaneous)
Q < Keq, the reaction proceeds in the forward direction spontaneously Q = Keq, the reaction is at equilibrium Q > Keq, the reaction favours the reverse direction.
7. where equilibrium constant K1 at temperature T1
equilibrium constnat K2 at temperature T2
8.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 3
PART AQUESTION 1
Using tabulated Thermodynamic data, To find: )Go, )Ho and )So at 298 K To verify: )Go = )Ho ! T )So
To determine: conditions of spontaneous reaction
(i) 4 HCl(g) + O2(g) ÷ 2 H2O(g) + 2 Cl2(g)
(a) )Go (reaction) = 3 n )Gof (products) ! 3 n )Go
In comparison to Part (a), )Go is not absolutely equal since we assumed )Ho
and )So to be temperature independent.
(c) Since )Go < 0 (i.e. !75.98 kJ), the reaction is spontaneous under standard conditions at298K (i.e. 25oC). According to )Go = )Ho ! T )So, the reaction is:
(1) favoured by negative )Ho (i.e. enthalpy driven, !114.40 kJ)(2) disfavoured by negative )So (!128.87 J K!1)
(d) )Go is negative and the reaction is spontaneous up to the temperature at which )Ho = T )So.ˆ T = )Ho ÷ )So = (!114.40 kJ) ÷ (!128.87 × 10!3 kJ K!1) = 887.7 K
Thus, the reaction is spontaneous at T < 888 K.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 4
(ii) H2(g) + Cl2(g) ÷ 2 HCl(g)
(a) )Go (reaction) = 3 n )Gof (products) ! 3 n )Go
(c) Since )Go < 0 (i.e. !190.6 kJ), the reaction is spontaneous under standard conditions at298 K (i.e. 25oC). According to )Go = )Ho ! T )So, the reaction is:
(1) favoured by negative )Ho (i.e. enthalpy driven, !184.62 kJ)(2) favoured by positive )So (i.e. entropy driven, +20.1 J K!1)
(d) Since )Go is always negative, the reaction occurs spontaneously at all temperatures.
(iii) C(graphite) ÷ C(diamond)
(a) )Go (reaction) = 3 n )Gof (products) ! 3 n )Go
f (reactants)= {)Go
f [C(diamond)]} ! {)Gof [C(graphite)]}
= {(2.900)} ! {(0)} kJ= +2.900 kJ
(b) )Ho (reaction) = 3 n )Hof (products) ! 3 n )Ho
f (reactants)= {)Ho
f [C(diamond)]} ! {)Hof [C(graphite)]}
= {(1.895)} ! {(0)} kJ= +1.895 kJ
)So (reaction) = 3 n )So (products) ! 3 n )So (reactants)
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 5
(a) a reaction with )Ho = +53.4 kJ and )So = +112.4 J K!1
Since both )Ho and )So are positive (i.e. unfavoured enthalpy and favoured entropy),the reaction is spontaneous above the temperature at which )Ho = T )So.
ˆ T > )Ho ÷ )So = (+53.4 kJ) ÷ (+112.4 × 10!3 kJ K!1) = 475 KThus, the reaction is spontaneous at T > 475 K.
(b) a reaction with )Ho = !29.4 kJ and )So = !91.2 J K!1
Since both )Ho and )So are negative (i.e. favoured enthalpy and unfavoured entropy),the reaction is spontaneous up to the temperature at which )Ho = T )So.
ˆ T < )Ho ÷ )So = (!29.4 kJ) ÷ (!91.2 × 10!3 kJ K!1) = 322 KThus, the reaction is spontaneous at T < 322 K.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 6
Î For a spontaneous reaction to occur, )Go must be negative.Ï Only reactions having like signs for both )Ho and )So exhibit temperature-
dependent spontaneity (‡ one term is unfavourable).)Go = )Ho ! T )So
when )Ho = !ve and )So = !veat low T: )Go = sve (spontaneous forward reaction)at high T: )Go = rve (nonspontaneous forward reaction)
when )Ho = +ve and )So = +veat low T: )Go = rve (nonspontaneous forward reaction) at high T: )Go = sve (spontaneous forward reaction)
PART AQUESTION 3
Wanted: a spontaneous reaction (i.e. )Go < 0) when a forward reaction occurs at low temperatures a reverse reaction occurs at high temperatures (i.e. a forward reaction occurs nonspontaneous at high temperatures)
To find: signs of )Ho and )So for such a reaction
(1) At low temperatures, the forward reaction proceeds spontaneously.ˆ )Go = !ve
It means )Ho term is dominant with the negative sign (i.e. favourable enthalpy change).
(2) At high temperatures, the forward reaction becomes nonspontaneous.It indicates sT )So term is dominant with the positive sign.(i.e. implying )S < 0, unfavourable entropy change).
ˆ Both )Ho and )So for such a reaction are negative.
PART AQUESTION 4
)Go depends on temperatures, but )Ho and )So don�t.
Strategy: Î Apply the Thermodynamics table to calculate )Ho(reaction) and )So(reaction) at 25oC. Ï Use the equation: )Go = )Ho ! T )So to calculate )Go at different temperatures. Ð Then use the equation: )Go = ! R T Rn Keq to calculate Kp at various temperatures.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 7
For a phase transition, )G = 0 (i.e. the (s) and (R) phases coexist in equilibrium)
From )G = )H ! T )S,
Assume )H and )S are independent of temperatures, )Hfus . )Horxn and )Sfus . )So
rxn
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 9
)Go depends on temperatures, but )Ho and )So don�t. Strategy: Î Use the Thermodynamic table to calculate )Ho(reaction) and )So(reaction) at 25oC. Ï Apply the equation: )Go = )Ho ! T )So to calculate )Go at 100oC. Ð Then use the equation: )Go = ! R T Rn Keq to calculate Kc at 100oC.
Note that: Kc = [H+] [OH!]
For comparison: At 25oC, Kc = Kw = 1.0 × 10!14 M2
At 100oC, Kc = 9.34 × 10!13 M2
PART AQUESTION 6
Given: H2O(R) ÷ H+(aq) + OH!(aq)From the Thermodynamic table at 25oC,)Ho
The reaction is:(1) favoured by negative )Ho (!23.0 kJ mol!1)(2) unfavoured by negative )So (!0.104 kJ K!1 mol!1)
The reaction occurs spontaneously up to the temperature at which )Ho = T )So.ˆ T < )Ho ÷ )So = (!23.0 kJ mol!1) ÷ (!0.104 kJ K!1 mol!1) = 221 K (!52oC)
Thus, the reaction is only spontaneous at T < 221 K, but not at T = 298 K (where )Go = +8.0 kJ mol!1).
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 11
)Go depends on temperatures, but )Ho and )So don�t.
ˆ )Go at 298 K will be somewhat different from )Go at 2000 K.
Strategy: Î Use the Thermodynamic table to calculate )Ho(reaction) and )So(reaction) at 25oC. Ï Apply the equation: )Go = )Ho ! T )So to calculate )Go at 2000K. Ð Then use the equation: )Go = ! R T Rn Keq to calculate Kp at 2000K.
The reaction H2(g) + ½ O2(g) ÷ H2O(R) is temperature-dependent because it is:(1) favoured (driven) by negative )Ho (!285.8 kJ mol!1)(2) unfavoured by negative )So (!163.2 J K!1 mol!1)
The reaction proceeds spontaneous up to the temperature at which )Ho = T )So
ˆ T = )Ho ÷ )So = (!285.8 kJ mol!1) ÷ (!163.2 × 10!3 kJ K!1 mol!1) = 1751 KAt 298 K (i.e. T < 1751 K), )Go = s237.2 kJ mol!1 (spontaneous reaction)At 2000 K (i.e. T > 1751 K), )Go = r40.6 kJ mol!1 (nonspontaneous reaction)
(c) )G at 2000 K when H2(g) and O2(g) are at 10.0 atm pressure each
Under non-standard state condition, p(H2) = p(O2) = 10.0 atm
Since )G is negative (i.e. !16.8 kJ mol!1), the reaction proceeds spontaneously from LEFTto RIGHT as written (i.e. the forward reaction is favourable).
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 13
(b) Since )G is negative (!17.6 kJ mol!1), the reaction is spontaneous under non-standard conditionsat 15oC when p(H2O,g) = 0.025 atm (i.e. the blue form is more stable).
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 14
PART AQUESTION
10
Given: AgCl(s) º Ag+(aq) + Cl!(aq)
From the Thermodynamic table,)Go
f [AgCl(s)] = !109.805 kJ mol!1
)Gof [Ag+(aq)] = 77.107 kJ mol!1
)Gof [Cl!(aq)] = !131.244 kJ mol!1
To find: Ksp and the solubility of AgCl(s) at 25oC
AgCl(s) º Ag+(aq) + Cl!(aq) (1:1 electrolyte)
)Go (reaction) = 3 n )Gof (products) ! 3 n )Go
f (reactants)= {)Go
f [Ag+(aq)] + )Gof [Cl!(aq)]} ! {)Go
f [AgCl(s)]}= {(77.107) + (!131.244)} ! {(!109.805)} kJ mol!1
This reaction is a phase transition process from H2O(R) to H2O(g) (i.e. vaporization).It indicates the reaction is at equilibrium, since the gas pressure equals the vapour pressure atthe temperature interest.
ˆ )G = 0
PART AMC #2
Given: For the reaction: HCO2H(R) ÷ HCO2H(g) at 298 K)Ho = 46.60 kJ)Go = 10.3 kJ)So = 122 J K!1 = 122 × 10!3 kJ K!1
To find: the normal boiling point of HCO2H(R)
At normal boiling point, the transition phase process from (R) to (g) is at equilibrium.Thus, )G = 0.
From )G = )H ! T )S,
Assume )Ho and )So are independent of temperatures (ˆ )Hvap . )Ho and )Svap . )So)
ˆ Tb.p. (HCO2H) = 382 K ! 273 K = 109oC
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 18
PART AMC #3
Given: Reaction studied: Hg(R) ÷ Hg(g)Kp = p(Hg,g) = 36.38 Pa (Note that Kp must be in terms of atm.)T = 100oC = (100 + 273) K = 373 K
To find: )Go of vaporization of Hg(R) at 100oC
Since 1 atm = 101.3 kPa = 101.3 × 103 Pa, 36.38 Pa = (36.38 Pa ÷ 101.3 × 103 Pa atm!1) = 3.5913 × 10!4 atm
Keq = Kp = p(Hg,g) = 3.5913 × 10!4 atm
At 100oC (i.e. 373 K), )Go = ! R T Rn Kp= ! (8.314 × 10!3 kJ mol!1 K!1) (373 K) Rn (3.591 × 10!4)= +24.59 kJ mol!1
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 24
Oxidation (loss of e!) occurs at the anode. Reduction (gain of e!) occurs at the cathode.
PART BQUESTION 1
Given: For the cell:Pt(s) , Fe2+(aq), Fe3+(aq) 1 Hg2+(aq) , Hg(R)
Î The anode is on the extreme left of the line (i.e. Pt(s)). Ï The cathode is on the extreme right (i.e. Hg(R)). Ð , is used to denote a change of physical state. Ñ 1 is used to denote the liquid junction.
(a) the anode half-reaction
Oxidation: Fe2+(aq) ÷ Fe3+(aq) + e! õoox = !0.771 V
2 e! are being transferred in the balanced redox reaction (ˆ n = 2)
ˆ [Cl!] = 7.352 × 10!3 M= 7.35 × 10!3 M
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 27
(1) The anode is Ni(s) and the cathode is Pt(s). (2) Ni(s) acts as reducing agent and oxidizes to Ni2+(aq) under oxidation.
Cl2(g) acts as oxidizing agent and reduces to Cl!(aq) under reduction. (3) Electrons flow from Ni(s) anode to the Pt(s) cathode.
There are two e! being transferred in the balanced redox reaction. (4) The anions (Cl!) migrate toward the Ni(s) anode via a salt bridge.
The cations (Ni2+) diffuse toward the Pt(s) cathode via a salt bridge. (5) Under standard conditions,
õocell is positive (+1.615 V), and the reaction is spontaneous.
Under non-standard state conditions,õ is also positive (+1.80 V), and the reaction occurs spontaneous as written.
(6) As the reaction proceeds, the Ni(s) anode loses mass and the concentrations of Ni2+(aq) andCl!(aq) increase in the anode and cathode compartments, respectively.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 28
PART BQUESTION 5
Given: For the reaction:AgCl(s) + Fe2+(aq) ÷ Ag(s) + Cl!(aq) + Fe3+(aq)
(a) a �shorthand� cell diagram for the reaction at an inert Pt anode
Î The anode (where oxidation occurs) is on the extreme left of the line (i.e. Pt(s)). Ï The cathode (where reduction occurs) is on the extreme right (i.e. Ag(s)). Ð , is used to separate the solid electrode from the liquid solution in contact. Ñ 1 is used to denote a salt bridge or porous barrier between the two solutions.
The cell diagram is: Pt(s) , Fe2+(aq), Fe3+(aq) 1 Cl!(aq) , AgCl(s) , Ag(s)
For the reduction of hydrogen electrode,2 H+(aq) + 2 e! ÷ H2(g) õo
red = 0
2 e! are being transferred in this reduction half-reaction. (ˆ n = 2)
ˆ õ = 0 ! (+0.366 V) = !0.366 V = !0.37 V (nonspontaneous)
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 31
Oxidation state of oxygen changes from zero in O2(g) to !2 in H2O(R) by gaining 2 e!. There are 7 oxygen atoms of initially elemental oxygen (i.e. 7/2 O2(g)), thus it requires
7× 2 e! = 14 e! to be transferred in the reaction.
PART BQUESTION 9
Given: A possible reaction for a fuel cell is:C2H6(g) + 7/2 O2(g) ÷ 2 CO2(g) + 3 H2O(R)
To find: )Go and õocell
)Go must be calculated from a tabulated Thermodynamic data using )Gof values.
To find: reagents that could reduce Cu2+(aq) to Cu(s),but not Al3+(aq) to Al(s)(from SRP Table)
Any reducing agent (via oxidation by losing e!) with SRP between +0.337 V and !1.662 V(i.e. between Cu(s) and Al(s)) can reduce Cu2+(aq) ÷ Cu(s), but cannot reduce Al3+(aq) ÷ Al(s).For examples, Pb(s), Ni(s), Cd(s), Fe(s), Cr(s), Zn(s), Mn(s) etc. can be the reducing agents forsuch a reaction.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 32
PART BQUESTION
11
Given: Ag(s) electrode (anode or cathode ?)Cu(s) electrode (anode or cathode ?)[AgNO3] = 0.50 M = [Ag+]õ = 0.62 VAs the cell works, [Cu2+] is increasing.
To find: initial [Cu2+]
Based on the statement �As the cell works, concentration of Cu2+(aq) is increasing.�,Cu metal must be dissolving as it is oxidized (i.e. Cu2+ ions go into the solution), and so Cu(s)
must be the anode.. Therefore, the cathode must be Ag(s).
The Overall Redox Reaction is: Cu(s) + 2 Ag+(aq) ÷ Cu2+(aq) + 2 Ag(s)
õocell = õo
ox + õored = (!0.337 V) + (+0.799 V) = +0.462 V
There are two electrons being transferred in the balanced redox reaction. (ˆ n = 2)
ˆ [Cu2+] = 1.125 × 10!6 M = 1.1 × 10!6 M
For the reduction of Ag+(aq) to Ag(s),Ag+(aq) + e! ÷ Ag(s) õo
red = +0.799 V2 Ag+(aq) + 2 e! ÷ 2 Ag(s) õo
red = +0.799 VThe SRP are the same in both cases. The charge on the silver electrode is related to the number of
electrons per surface area of silver. Thus, the ration of e! to Ag atom in both cases is the same.
Therefore, electrode potentials are intensive properties and do not depend on the size of the electrode or themanner in which the half-reaction is balanced.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 33
In order to obtain the Ksp [AgBr(s)] expression, both Ag+(aq) and Br!(aq) must be on the�product� side. Therefore, rearrange the above two equations as follows:
!1× Î Ag(s) ÷ Ag+(aq) + e! õoox = !0.799 V (i.e. oxidation)
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 35
PART BQUESTION
14
CATHODE is an electrode where reduction takes place (i.e. gaining e!). ANODE is an electrode where oxidation takes place (i.e. losing e!).
(a) NaBr Given: aqueous solution of NaBr(aq)(i.e. Na+(aq), Br!(aq) and H2O(R))
To write: cathode and anode reactions
CATHODE
In aqueous solution, the cation (Na+) is not the only species that can be reduced. H2O(R) canbe reduced as well. The two possible reduction reactions are:
‡ H2O(R) has a higher reduction potential (i.e. more +ve) than Na+(aq).ˆ H2O(R) is more readily reduced than Na+(aq)
andH2(g) is liberated at the cathode.
ANODE
In aqueous solution, the anion (BrG) is not the only species that can be oxidized. H2O(R) can beoxidized as well. The two possible oxidation reactions are:
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 36
PART BQUESTION
14
(b) AgF Given: aqueous solution of AgF(aq)(i.e. Ag+(aq), F!(aq) and H2O(R))
To write: cathode and anode reactions
CATHODE
In aqueous solution, the cation (Ag+) is not the only species that can be reduced. H2O(R) canbe reduced as well. The two possible reduction reactions are:
‡ Ag+(aq) has a higher reduction potential (i.e. more +ve) than H2O(R).ˆ Ag+(aq) is more readily reduced than H2O(R)
andAg(s) is formed at the cathode.
ANODE
In aqueous solution, the anion (FG) is not the only species that can be oxidized. H2O(R) can beoxidized as well. The two possible oxidation reactions are:
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 37
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 38
PART BQUESTION
14
(c) NaF Given: aqueous solution of NaF(aq)(i.e. Na+(aq), F!(aq) and H2O(R))
To write: cathode and anode reactions
CATHODE
In aqueous solution, the cation (Na+) is not the only species that can be reduced. H2O(R) canbe reduced as well. The two possible reduction reactions are:
‡ H2O(R) has a higher reduction potential (i.e. more +ve) than Na+(aq)ˆ H2O(R) is more readily reduced than Na+(aq)
andH2(g) is liberated at the cathode.
ANODE
In aqueous solution, the anion (FG) is not the only species that can be oxidized. H2O(R) can beoxidized as well. The two possible oxidation reactions are:
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 39
PART BQUESTION
15
Given: aqueous solution of NaOH(i.e. Na+(aq), OH!(aq) and H2O(R))
To find: products in each electrode To write: the overall reaction
CATHODE is an electrode where reduction takes place (i.e. gaining e!).ANODE is an electrode where oxidation takes place (i.e. losing e!).
CATHODE
In aqueous solution, the cation (Na+) is not the only species that can be reduced. H2O(R) canbe reduced as well. The two possible reduction reactions are:
‡ H2O(R) has a higher reduction potential (i.e. more +ve) than Na+(aq).ˆ H2O(R) is more readily reduced than Na+(aq)
andH2(g) is liberated at the cathode.
ANODE
In aqueous solution, the anion (OHG) is not the only species that can be oxidized. H2O(R) canbe oxidized as well. The two possible oxidation reactions are:
Î 4 OH!(aq) ÷ O2(g) + 2 H2O(R) + 4 e! õoox = !0.401 V
(Note that K+(aq) and Na+(aq) are spectator ions in aqueous solution.)
Ï The oxidation state of chlorine increases from !1 in Cl!(aq) to zero in Cl2(g). Therefore, Cl islosing one e! in this half-reaction (i.e. oxidation), and so it is oxidized.
(Note that there are 6 e! being transferred in the redox reaction.)
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 42
Oxidation = loss of e!
Reduction = gain of e!
H2S(aq): acts as reducing agent which undergoes oxidation. MnO4
!(aq): acts as oxidizing agent which undergoes reduction.
PART BMC #4
Given: For the balanced redox reaction:5 H2S(aq) + 2 MnO4
!(aq) + 6 H+(aq) ÷ 2 Mn2+(aq) + 5 S(s) + 8 H2O(R)
To find: the number of electrons transferred
Î OxidationThe oxidation state of sulfur increases from !2 in H2S(aq) to zero in S(s). Therefore, S
atom is losing 2 e! in this half-reaction (i.e. oxidation), and so it is oxidized.H2S(aq) ÷ S(s)H2S(aq) ÷ S(s) + 2 H+(aq)H2S(aq) ÷ S(s) + 2 H+(aq) + 2 e!
Ï ReductionThe oxidation state of manganese decreases from +7 in MnO4
!(aq) to +2 in Mn2+(aq).Therefore, Mn atom is gaining 5 e! in this half-reaction (i.e. reduction), and so it is reduced.
MnO4!(aq) ÷ Mn2+(aq)
MnO4!(aq) ÷ Mn2+(aq) + 4 H2O(R)
MnO4!(aq) + 8 H+(aq) ÷ Mn2+(aq) + 4 H2O(R)
MnO4!(aq) + 8 H+(aq) + 5 e! ÷ Mn2+(aq) + 4 H2O(R)
For the overall balanced redox reaction: 5× Î 5 H2S(aq) ÷ 5 S(s) + 10 H+(aq) + 10 e!
ˆ There are 10 e! being transferred in the redox reaction.
ALTERNATE EXPLANATION:Each sulfur atom loses 2 e!, thus it requires 10 e! transferred for 5 sulfur atoms in the redox
reaction. On the other hand, each manganese atom gains 5 e!, thus it also requires 10 e! transferredfor 2 Mn atoms in the redox reaction.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 43
PART BMC #5
Given: C6H12O6(s) + 6 O2(g) ÷ 6 CO2(g) + 6 H2O(R)
To find: the number of electrons transferred
Î Oxidation state of oxygen decreases from zero in O2(g) to !2 in H2O(R) by gaining 2 e!.There are 12 oxygen atoms of initially elemental oxygen (i.e. 6 O2(g)), thus it requires:
12× 2 e! = 24 e!
Ï Oxidation state of carbon increases from zero in C6H12O6(s) to +4 in CO2(g) by losing 4 e!.There are 6 carbon atoms in C6H12O6(s), thus it requires:
6× 4 e! = 24 e!
ˆ There are 24 e! being transferred in the reaction.
CHEM*131 (W 02) REVIEW QUESTIONS FOR FINAL EXAM PAGE - 45
PART BMC #8
Given: 1.0 M aqueous solution of KI(aq)(i.e. K+(aq), I!(aq) and H2O(R))
To find: product(s) formed at the anode
ANODE is the electrode where oxidation takes place (i.e. losing e!).
ANODE
In aqueous solution, the anion (IG) is not the only species that can be oxidized. H2O(R) can beoxidized as well. The two possible oxidation reactions are:
‡ I!(aq) has a higher oxidation potential (i.e. more +ve) than H2O(R).ˆ I!(aq) is more readily oxidized than H2O(R)
andI2(s) is produced at the anode.
CATHODE
In aqueous solution, the cation (K+) is not the only species that can be reduced. H2O(R) can bereduced as well. The two possible reduction reactions are:
1 mole of Fe3+(aq) deposits 1 mole of Fe(s)ˆ n(Fe) = n(Fe3+) = 1.25 × 10!1 mol
Ïcharge (Q) = current (i) × time (t)
ˆ Q = (25.0 A) × (1800 s) = 4.50 × 104 C
n(e!) = Q ÷ ö= (4.50 × 104 C) ÷ (96500 C / mol e!)= 4.663 × 10!1 mol e!
It requires 3 moles of e! to plate out 1 mole of Fe(s)n(Fe) deposited = n(e!) × (1 mole Fe(s) ÷ 3 moles e!)
= (4.663 × 10!1 mol) × (a)= 1.554 × 10!1 mol
ˆ Fe(NO3)3 solution is the limiting reagent, not the time and current.(It takes only -24 min with a constant current of 25.0 A to completely deposit Fe(s) from 250 mLof a 0.500 M Fe(NO3)3 solution. Verify the calculated time yourself!)
M(Fe) = 55.847 g mol!1
ˆ mass(Fe) deposited = n(Fe) × M(Fe)= (1.25 × 10!1 mol) × (55.847 g mol!1)= 6.9809 g= 6.98 g